While working on a homework problem, I am baffled by the following statement:
Let $(X,\mathcal{M},\mu)$ be a finite measure space, $\mathcal{N}$ a sub-$\sigma$-algebra of $\mathcal{M}$, and $\nu = \mu|_{\mathcal{N}}$. If $f \in L^1(\mu)$, there exists $g\in L^1(\nu)$ such that $\int_E f\, d\mu = \int_E g\, d\nu$ for all $E \in N$; if $g'$ is another such function then $g = g'$ $\nu$-a.e.
My question: how is $g$ not just $f$?
I have a proof of the proposition; however it is mechanical and elides any sense of intuition.
My issues:
- Since $E \in \mathcal{N}$ and $\mathcal{N} \subset \mathcal{M}$, then we have $E \in \mathcal{M}$.
- Since $f$ is $\mathcal{M}$-measurable, is it not $\mathcal{N}$-measurable?
In the event that $f$ is $\mathcal{N}$-measurable, then in fact $g=f$. However, $f$ need not be $\mathcal{N}$-measurable, whereas the function $g$ must be. Such a function would have the property that $g^{-1}(E) \in \mathcal{N}$ for any $E$ in the $\sigma$-algebra of the range of $g$.